一道微积分 (y^2(1-x^2))^(1/2)dy=arcsinxdx,y(0)=1
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一道微积分 (y^2(1-x^2))^(1/2)dy=arcsinxdx,y(0)=1
一道微积分 (y^2(1-x^2))^(1/2)dy=arcsinxdx,y(0)=1
一道微积分 (y^2(1-x^2))^(1/2)dy=arcsinxdx,y(0)=1
(y^2(1-x^2))^(1/2)dy=arcsinxdx 分离变量
=> ydy = arcsinx *(1-x^2)^(-1/2) dx
=> ∫ ydy = ∫ arcsinx *(1-x^2)^(-1/2) dx 积分
=> y^2 /2 = (1/2) (arcsinx)^2 + C1
=> y^2 = (arcsinx)^2 + C
求解微分方程:y²[√(1-x²)]dy=(arcsinx)dx,,y(0)=1
分离变量:y²dy=[(arcinx)/√(1-x²)]dx...........(1)
积分之,得y³/3=∫[(arcinx)/√(1-x²)]dx
令arcsinx=u,则x=sinu,dx=cosudu,故∫[(arcinx)...
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求解微分方程:y²[√(1-x²)]dy=(arcsinx)dx,,y(0)=1
分离变量:y²dy=[(arcinx)/√(1-x²)]dx...........(1)
积分之,得y³/3=∫[(arcinx)/√(1-x²)]dx
令arcsinx=u,则x=sinu,dx=cosudu,故∫[(arcinx)/√(1-x²)]dx=∫ucosudu/√(1-sin²u)=∫udu=u²/2
=(arcsinx)²/2,代入(1)式得y³/3=(arcsinx)²/2+C/3
即得通y³=(3/2)(arcsinx)²+C,y(0)=[(3/2)(arcsin0)²+C]^(1/3)=C^(1/3)=1,故C=1
于是得方程的特解为 y³=(3/2)(arcsinx)²+1.
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